\subsection{Fourier tranform}
Momentum operator
\begin{equation}
\mathbf{p}=-i\hbar\nabla
\end{equation}
So 
\begin{equation}
\avt{\vr}{\vk}=\nth{\sqrt{V}}e^{i\vk\cdot\vr}
\end{equation}
\begin{equation}
\avs{\vk}{\hat{O}}{\vp}=O(\vk-\vp)
\end{equation}

In continuous limit
\begin{equation}
f(\omega)=\nth{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(x)}e^{-i\omega{x}}dx
\end{equation}
\begin{equation}
f(x)=\nth{\sqrt{2\pi}}\int_{-\infty}^{\infty}{f(\omega)}e^{i\omega{x}}d\omega
\end{equation}

\subsection{$\delta$ related}
\begin{equation}
\lim_{\eta\rightarrow0^{+}}\nth{x\pm{}i\eta}=\mp{}i\pi\delta{(x)}+P\,\nth{x}
\end{equation}


\subsection{Fermi gas}
Fermi momentum for single-species density
\begin{equation}
k_{F}=(6\pi{}n)^{1/3}
\end{equation}
Typical Fermi gas experiments, density is $n:10^{18}\sim 10^{21}\text{m}^{-3}$, $k_{F}:2.7\times10^{6}\sim2.7\times10^{8}m^{-1}$.
${}^{6}\text{Li}$ Fermi energy is
\begin{align}
E_{F}({}^{6}\text{Li})&=\frac{\hbar^{2}k_{F}^{2}}{2m}\\
&=3.9\times10^{-30}\sim3.9\times10^{-24}J\\
&=2.4\times10^{-11}\sim2.4\times10^{-5}eV\\
&=5.9\times10^{3}\sim5.9\times10^{9}Hz\cdot{}h\\
&=2.8\times10^{-7}\sim2.8\times10^{-1}T\cdot{}k_{B}\\
&=4.2\times10^{-3}\sim4.2\times10^{3}Gauss\cdot{}\mu_{B}
\end{align}
$\mu_{B}=e\hbar/2m_{E}\approx927\times10^{-26}JT^{-1}$ is the electronic meganetic moment.
\begin{equation}
a_{0}=n^{-1/3}=10^{-7}\sim10^{-6}m^{-1}=100\sim1000\AA
\end{equation}
